# Spectrum of the Resolvent of a Self-Adjoint Operator

Let $\mathcal{H}$ be a Hilbert space, and $A$ a self-adjoint operator with domain $D_{A} \subseteq \mathcal{H}$. Assume that $\lambda_0 \in \rho(A)$, where $\rho(A)$ is the resolvent set of $A$. For any $z \in \rho(A)$, let $R_{A}(z)=(A - z I)^{-1}$ be the resolvent of $A$.

Choose $\lambda \neq \lambda_0$. Then it is well known that $\lambda \in \rho(A)$ if and only if $(\lambda - \lambda_0)^{-1} \in \rho(R_{A}(\lambda_0))$ (see e.g. Schmudgen, Unbounded Self-adjoint operators on Hilbert Space, Proposition 2.10). So we have (note that by the spectral theorem $\sigma(A)$ is nonempty): $$\sigma(R_{A}(\lambda_0)) \backslash \{0\} = \left \{ \frac{1}{\mu - \lambda_0} : \mu \in \sigma(A) \right \}.$$ If $A$ is a bounded operator on $\mathcal{H}$, then $0 \in \rho(R_{A}(\lambda_0))$, so that in this case, being $\sigma(A)$ closed, we have $$\sigma(R_{A}(\lambda_0)) = \left \{ \frac{1}{\mu - \lambda_0} : \mu \in \sigma(A) \right \} = \text{closure} \left \{ \frac{1}{\mu - \lambda_0} : \mu \in \sigma(A) \right \}.$$ Now suppose that $A$ is unbounded. In this case $0 \in \sigma(R_{A}(\lambda_0))$. If we could prove that $0$ is not an isolated point of $\sigma(R_{A}(\lambda_0))$ (which is the same to say that $\sigma(A)$ is not bounded), we could conclude also in this case that $$\sigma(R_{A}(\lambda_0)) = \text{closure} \left \{ \frac{1}{\mu - \lambda_0} : \mu \in \sigma(A) \right \}.$$ So my question is the following: if $A$ is unbounded, can $\sigma(A)$ be bounded?

PS This question arouse from the answer given by TrialAndError in this post Norm of the Resolvent

• What is $R_A(\lambda_0)$? Is it $(A-\lambda_0)^{-1}$? I think you also have a typo in your first paragraph. Should it be $(\lambda-\lambda_0)^{-1}$? – Cameron Williams Jul 11 '16 at 15:15
• $R_A(\mu)$ is usually the inverse of $\mu I - A$ – supinf Jul 11 '16 at 15:16
• This question might be of interest to you: math.stackexchange.com/questions/194681/… – Cameron Williams Jul 11 '16 at 15:21
• Dear Cameron, the last answer to the post you quoted is actually related to the argument given by TrialAndError in one of his comments math.stackexchange.com/questions/1855151/norm-of-the-resolvent He actually says that for any normal operator $N$ with a bounded spectrum , we have $N = - \frac{1}{2 \pi i} \int_{C_{R}} \lambda R_{N}( \lambda) d \lambda$, where $C_R=\{ \lambda \in \mathbb{C}: |\lambda| = R \}$, and $R > 0$ is big enough so that $C_R$ encircles the spectrum of $N$. But I can't see how to prove this representation – Maurizio Barbato Jul 11 '16 at 16:00
• If we denote by $\mathcal{B(H)}$ the Banach space of bounded operators on $\mathcal{H}$, then $R_{N}: \rho(N) \rightarrow \mathcal{B(H)}$ is an analytic function. So the integral $\oint_{C_R} \lambda R_{N}(\lambda) d \lambda$ is well defined and by definition is in $\mathcal{B(H)}$. So if we could prove the above representation for $N$, then we could conclude that $N$ cannot be unbounded if it has a bounded spectrum. – Maurizio Barbato Jul 11 '16 at 16:06

Take $r > \max \sigma(A)$. Then $R_A(r)$ is self-adjoint and bounded. If $0$ is not in its spectrum, then $A = (R_A(r)^{-1}+rI$ is bounded. If $0$ is in its spectrum, it is an isolated point of the spectrum and therefore must be an eigenvalue: $R_A v = 0$ for some $v \in \mathcal H$. But that is impossible since $R_A(r) = (A-rI)^{-1}$, i.e. $R_A(r) v = u$ where $u \in D_A$ and $(A-rI) u = v$.
You mentioned you like Complex Analysis. So I thought I'd offer a proof using Complex Analysis applied to the resolvent. The proof comes down to evaluating the integral around all finite singularities of $(\lambda I-A)^{-1}x$ by determining the residue at $\infty$, which turns out to be $x$. This equivalence forces the completeness of spectral expansions for normal operators. It's a type of Complex Analysis conservation law that allows you to expand $x$ in terms of integrals in the finite plane. If the singularities are all discrete in the finite plane, you end up with an eigenfunction expansion of $x$. Continuous spectrum can lead to integral expansions, such as the classical Fourier integral expansions. More generally, the Spectral Theorem for sefadjoint operators can be proved using this conversation law; completeness is established by knowing that the resiude at infinity of $(\lambda I-A)^{-1}$ is $I$. So the technique is worth learning.
Suppose $A$ is a closed densely-defined normal operator on a Complex Hilbert space $\mathcal{H}$, and suppose that $\sigma(A)$ is a bounded set. By the previous problem you referenced, $$\|(\lambda I -A)^{-1}\| \le \frac{1}{\mbox{dist}(\lambda,\sigma(A))}.$$ Therefore, $\lim_{\lambda\rightarrow\infty} (\lambda I-A)^{-1}=0$, and, for a fixed $x\in\mathcal{D}(A)$, the following limit is uniform in $\lambda$: $$\lim_{\lambda\rightarrow\infty}\lambda(\lambda I-A)^{-1}x=\lim_{\lambda\rightarrow\infty}x+(\lambda I-A)^{-1}Ax = x.$$ If $x\in\mathcal{D}(A)$, and if $R$ is large enough that $\sigma(A)\subseteq \{ \lambda : |\lambda| < R \}$, then $$\frac{1}{2\pi i}\oint_{|\lambda|=R}(\lambda I-A)^{-1}xd\lambda = \lim_{R\rightarrow\infty}\frac{1}{2\pi i}\oint_{|\lambda|=R}\lambda(\lambda I-A)^{-1}x\frac{d\lambda}{\lambda}=x.$$ Because $\mathcal{D}(A)$ is dense and $\oint_{|\lambda|=R}(\lambda I-A)^{-1}d\lambda$ is a bounded operator, then $$\frac{1}{2\pi i}\oint_{|\lambda|=R}(\lambda I-A)^{-1}d\lambda =I.$$ For $x\in\mathcal{D}(A)$, \begin{align} Ax & = \frac{1}{2\pi i}\oint_{|\lambda|=R}(\lambda I-A)^{-1}Ax\,d\lambda \\ & = \frac{1}{2\pi i}\oint_{|\lambda|=R}-x+\lambda(\lambda I-A)^{-1}x\,d\lambda \\ & = \left(\frac{1}{2\pi i}\oint_{|\lambda|=R}\lambda (\lambda I-A)^{-1}d\lambda\right)x \end{align} So $A$ is bounded on $\mathcal{D}(A)$, which also forces $\mathcal{D}(A)=\mathcal{H}$ because $A$ is closed.